Problem: The polynomial $p(x)=x^3-19x-30$ has a known factor of $(x+2)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
We know $(x+2)$ is a factor of $p(x)$. This means that $p(x)=(x+2)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x+2)$ Notice that $p(x)$ is missing a $2^{\text{nd}}$ degree term. Let's add it as $0x^2$. $\begin{array}{r} x^2-\phantom{1}2x-15 \\ x+2|\overline{x^3+0x^2-19x-30} \\ \mathllap{-(}\underline{x^3+2x^2\phantom{-19x-30}\rlap)} \\ -2x^2-19x-30 \\ \mathllap{-(}\underline{-2x^2-\phantom{1}4x\phantom{-30}\rlap)} \\ -15x-30 \\ \mathllap{-(}\underline{-15x-30\rlap)} \\ 0 \end{array}$ We find that $q(x)=x^2-2x-15$. Factoring $q(x)$ We can factor $x^2{-2}x{-15}$ as $(x+m)(x+n)$ where $m+n={-2}$ and $m\cdot n={-15}$. Such numbers are $3$ and $-5$, so the factored expression is $(x+3)(x-5)$. Putting it all together $\begin{aligned} p(x)&=x^3-19x-30 \\\\ &=(x+2)(x^2-2x-15) \\\\ &=(x+2)(x+3)(x-5) \end{aligned}$